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\noindent
Important example: Surface of revolution again! We
take $\eta(u) = (f(u), 0, g(u))$ in the $xz$-plane
and rotate about the $z$-axis. ($\eta : [a, b] \to
\RR^3$ smooth, injective, $\eta' \neq 0$, $f >
0$).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/839903c6b43411ed.png}
\end{center}
Take the usual $\sigma$:
\[ \sigma(u, v) = (f(u) \cos v, f(u) \sin v, g(u)) \]
$a < u < b$, $v \in (0, 2\pi)$. The FFF is
\[ (f'^2 + g'^2) \dd u^2 + f^2 \dd v^2 \]
WLOG we assume $\eta$ is parametrised by
arc-length so the FFF becomes
\[ \dd u^2 + f^2 \dd v^2 \]
Then Lagrangian for the geodesics is
\[ L = \half (\dot{u}^2 + f^2 \dot{v}^2) \]
(E-L) equations implies $\pfrac{L}{u} = ff'
\dot{v}^2$
\[ \pfrac{L}{\dot{u}} = \dot{u} \implies
\boxed{\ddot{u} = ff' \dot{v}^2} \]
\[ \boxed{\dfrac{}{t} \left( \pfrac{L}{\dot{q}_i}
\right) = \pfrac{L}{q_i}} \]
\[ \pfrac{L}{v} = 0 \qquad \pfrac{L}{\dot{v}} =
f^2 \dot{v} \]
\[ (*) \begin{cases}
    \dfrac{}{t} (f^2 \dot{v}) = 0 \\
    \ddot{u} = ff' \dot{v}^2
\end{cases} \]
We also know that geodesics travel with constant
speed:
\[ \begin{cases}
    \dot{u}^2 + f^2 \dot{v}^2 = \text{const} \\
    f^2\dot{v} = c
\end{cases} \]

\begin{example*}[``completely integrable'']
    Meridians: $v = v_0$m if $u(t) = t + u_0$, $t
    \mapsto (t + u_0, v_0)$ is a geodesic with
    speed 1 through $(u_0, v_0)$ (check this
    works).
    \myskip
    Parallels: $u = u_0$, $\dot{v} = a
    \mid_{f(u_0)}$. From ($*$) we see that we need
    \[ \boxed{f'(u_0) = 0} \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/70085378b43611ed.png}
    \end{center}
    Let's look at the conserved quantity $f^2
    \dot{v}$ in more detail:
    .image
    Suppose $\gamma$ makes an angle $\theta$ with
    a parallel of radius $\rho = f$. Write as
    usual $\gamma = \sigma(u(t), v(t))$,
    $\dot{\gamma} = \sigma_u \dot{u} + \sigma_v
    \dot{v}$ and note that $\sigma_v$ is tangent
    to the parallel since $\sigma_v = (-f \sin v,
    f \cos v, 0)$. Thus
    \[ \cos\theta = \frac{\langle \sigma_v,
    \sigma_u \dot{u} + \sigma_v \dot{v}
    \rangle}{|\sigma_v| |\dot{\gamma}|} \]
    Assume $\gamma$ is parametrised by arc-length,
    so $|\dot{\gamma}| = 1$. Using that $F = 0$
    and $G = f^2$ we get
    \[ \cos\theta = \frac{f^2 \dot{v}}{f} =
    f\dot{v} \]
    and therefore if $\gamma$ is geodesic,
    \[ \boxed{\rho \cos\theta = \text{constant}} \]
    \emph{Clairut's relation}. This is just
    another way to write the conservation law
    arising from $\pfrac{L}{v} = 0$.
\end{example*}

\begin{example*}
    Sub-example: ellipsoid of revolution.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/bf4affecb43811ed.png}
    \end{center}
    $\rho \cos\theta = c$, $c = \rho_0
    \cos\theta_0 > 0$, $\theta_0 \in [0, \pi/2)$.
    $c = \rho\cos\theta \le \rho$. This means that
    $\gamma$ must move between the region bounded
    by the parallels of radius $c$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/e5f1994eb43811ed.png}
    \end{center}
\end{example*}

Recall Picard's theorem for ODEs: $I = [t_0 - a,
t_0 + a] \subset \RR$. $B = \{x : \|x - x_0\| \le
b\} \subseteq \RR^n$. $f : I \times B \to \RR^n$
and Lipschitz in the second variable
\[ \|f(t, x_1) - f(t, x_2)\| \le K\|x_1 - x_2\| \]
Then
\[ \begin{cases}
    \dfrac{x(t)}{t} = f(t, x(t)) \\
    x(t_0) = x_0
\end{cases} \]
has a unique solution for some interval $|t - t_0|
< h$.

\myskip
Addendum: If $f$ is smooth, then the solution is
smooth \emph{and} depends smoothly on the initial
condition. In our setting we have:
\begin{align*}
    \dfrac{}{t} (E\dot{u} + F \dot{v})
    &= \half (E_u \dot{u}^2 + 2F_u \dot{u} \dot{v}
    + G_u \dot{v}^2) \\
    \dfrac{}{t} (F\dot{u} + G \dot{v})
    &= \half (E_v \dot{u}^2 + 2F_v \dot{u} \dot{v}
    + G_v \dot{v}^2)
    \begin{pmatrix}
        E & F \\
        F & G
    \end{pmatrix}
    \begin{pmatrix}
        \ddot{u} \\
        \ddot{v}
    \end{pmatrix}
    &=
    \mathcal{A} (u, \sigma, \dot{u}, \dot{v})
\end{align*}
Since the matrix is invertible, we can write the
geodesics equations as:
\begin{align*}
    \ddot{u}
    &= A(u, v, \dot{u}, \dot{v}) \\
    \ddot{v}
    &= B(u, v, \dot{u}, \dot{v})
\end{align*}
for smooth $A$ and $B$. We can turn this into a
first order system by the usual trick:
\[ \dot{u} = X, \dot{v} = Y \]
Then
\[ \begin{cases}
    \dot{u} = X \\
    \dot{v} = Y \\
    \dot{X} = A(u, v, X, Y) \\
    \dot{Y} = B(u, v, X, Y)
\end{cases} \]
So Picard's theorem applies, noting that since $A$
and $B$ are smooth, a local bound on $\|DA\|$ and
$\|DB\|$ wil give the Lipschitz conditions. So we
get:

\begin{corollary}
    Let $\Sigma$ be a smooth surface in $\RR^3$.
    For $p \in \Sigma$ and $v \in T_p\Sigma$,
    there is $\eps > 0$ and a unique geodesic
    $\gamma : (-\eps, \eps) \to \Sigma$ with
    initial conditions $\gamma(0) = p$ and
    $\dot{\gamma}(0) = v$. Moreover, $\gamma$
    depends smoothly on $(p, v)$.
\end{corollary}

\begin{hiddenflashcard}[Picard-geodesics]
Picard geodesics? \\
\cloze{
For $p \in \Sigma$ and $v \in T_p\Sigma$, there is
$\eps > 0$ and a unique geodesic $\gamma : (-\eps,
\eps) \to \Sigma$ with initial conditions
$\gamma(0) = p$ and $\dot{\gamma}(0) = v$. Moreover,
$\gamma$ depends smoothly on $(p, v)$.}
\end{hiddenflashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/31ce52fcb43a11ed.png}
\end{center}